Minibus stochastic process

Recently I had a talk in Pereslavl',
so I needed to get up early and travel to bus station, because the only way to get to Pereslavl' on public transport
is by intercity bus.

That was a long way (about 6 hours one direction). But first step was to get to the nearest underground station,
which is not so trivial in mornings.
People hurry, traffic jams make public transport circulate slowly, and as a result,
there is crowding in nearly every bus/minibus moving towards city center or nearest underground station.

After I found a place for me in one of minibuses I was quite sure that there is at most free place for one or two
passengers.

Of course, I was wrong.
The minibus was moving its way and driver skipped the stops if nobody wanted to get off, since minibus was already
filled.
But when he stopped and one of passengers got off, two new were coming into, which made it even more crowdy inside.

Each time I was sure, that after looking inside people
won't enter the minibus, and each time they did. Exactly two new passengers, others didn't dare.

So I came up with such a simple stochastic process:
At $t=0$ there is only one passenger.
During small time interval $\Delta t$ each passenger with probability $\alpha \Delta t$ wants to get off.
When he gets off, two new passengers are entering the bus.

Let's
introduce an analytical over $s$ function that will describe the process:$F(t, s) = \sum_{n=0}^{+\infty} p_n
(t) s^n$, where $p_n(t)$ is probability that at the moment $t$ there are exactly $n$ passengers.

Note:

$F(t,1) = 1$ for any $t$, since the probabilities sum up to 1.

$F(0,s) = s$, since initially there was exactly one passenger

Now let's think about how this system evolves. During small time interval $\Delta t$ with probability $\alpha n
\Delta t$ the number of passengers is incremented (by one).